The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are W1 = 54.5 N, W2 = 32.7 N, and W3 = 29.2 N. Determine the magnitude of the normal force that the table exerts on box 1.
To determine the magnitude of the normal force that the table exerts on box 1, we need to analyze the forces acting on box 1.
First, let's consider the forces acting vertically. The weight of box 1 (W1 = 54.5 N) acts downwards, while the normal force (N1) exerted by the table acts upwards. These two forces must be equal in magnitude and opposite in direction according to Newton's third law.
Next, let's consider the forces acting horizontally. Since box 1 is at rest and there is no acceleration horizontally, the net force in the horizontal direction must be zero. The tension in the rope (T) pulling box 1 to the right and the friction force between box 1 and the table (Ffr) pulling box 1 to the left are the only horizontal forces.
Since the pulley is frictionless and massless, the tension in the rope is the same on both sides of the pulley. This means that the tension in the rope pulling box 1 (T1) is equal in magnitude to the tension in the rope pulling box 2 (T2). However, the directions of these tensions are different. T1 pulls to the right, while T2 pulls to the left.
Since box 1 and box 2 are in contact with each other, they experience an equal and opposite contact force. The tension in the rope between box 1 and box 2 (T2) creates a downward force on box 1, pushing it against the table. This contact force is directly related to the normal force (N1) exerted by the table on box 1.
Therefore, the magnitude of the normal force (N1) exerted by the table on box 1 is equal to the magnitude of the contact force between box 1 and box 2, which is equal to the tension in the rope between box 1 and box 2 (T2).
To calculate T2, we need to consider the forces acting on box 2. Box 2 has two forces acting vertically: its weight (W2 = 32.7 N) acting downwards and the tension in the rope between box 2 and box 1 (T2) acting upwards. Since box 2 is at rest in the vertical direction, the net force in the vertical direction must be zero. Therefore, the sum of these two forces must be equal in magnitude.
W2 - T2 = 0
T2 = W2
Thus, the tension in the rope between box 1 and box 2 (T2) is equal to the weight of box 2 (W2 = 32.7 N).
Since the magnitude of the normal force (N1) exerted by the table on box 1 is equal to the magnitude of T2, we can conclude that the normal force exerted by the table on box 1 is 32.7 N.